The main supply voltage to a room is 120 V. The resistance of the lead wires is $6\Omega .$ A 60 W bulb is already giving light. What is the decrease in voltage across the bulb when a 240 W heater is switched on?

$R_{60}=\frac{120\times 120}{60}\Omega =240\Omega$ 

Current $=\frac{120}{240+6}A=\frac{120}{246}A$

Voltage across bulb $=\frac{120}{246}\times 240V=117.1V$

$R_{240}=\frac{120\times 120}{240}\Omega =60\Omega$

Resistance of parallel combination is

$\frac{60\times 240}{60+240}\Omega =48\Omega$

Total resistance is $(48+6)\Omega =54\Omega .$

Current I is 120/54 A. Voltage across parallel combination is$\frac{120}{54}\times 48V=106.7V$

Change in voltage is $(117.1-106.7)V=10.4V$